Answer:
5.714 liters of the 5% solution and 4.286 of the 40%.
Step-by-step explanation:
Let x be the volume of 40% solution and y = volume of the 5% solution.
x + y = 10
0.40x + 0.05y = 0.20(x + y)
From the first equation x = 10 -y so we have:
0.40(10 - y) + 0.05y = 0.20( 10 - y + y)
4 - 0.40y + 0.05y = 2
-0.35y = -2
y = 5.714 liters of the 5%.
and x = 10 - 5.714 = 4.286 liters of the 40% solution.
Step-by-step explanation:
Answer:
11 sq units
Step-by-step explanation:
(6×4) - ½(6×1 + 4×3 + 2×4)
= 24 - ½(6+12+8)
= 24 - ½(26)
= 24-13
= 11 sq units
Answer:
H0: μ = 5 versus Ha: μ < 5.
Step-by-step explanation:
Given:
μ = true average radioactivity level(picocuries per liter)
5 pCi/L = dividing line between safe and unsafe water
The recommended test here is to test the null hypothesis, H0: μ = 5 against the alternative hypothesis Ha: μ < 5.
A type I error, is an error where the null hypothesis, H0 is rejected when it is true.
We know type I error can be controlled, so safer option which is to test H0: μ = 5 vs Ha: μ < 5 is recommended.
Here, a type I error involves declaring the water is safe when it is not safe. A test which ensures that this error is highly unlikely is desirable because this is a very serious error. We prefer that the most serious error be a type I error because it can be explicitly controlled.
You just do 12•.65. Then that’s your answer.
